leetcode -- Populating Next Right Pointers in Each Node

Given a binary tree

    struct TreeLinkNode {
      TreeLinkNode *left;
      TreeLinkNode *right;
      TreeLinkNode *next;
    }

 Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set toNULL.

Initially, all next pointers are set to NULL.

Note:

• You may only use constant extra space.
• You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

 For example,

Given the following perfect binary tree,

         1
       /  \
      2    3
     / \  / \
    4  5  6  7

 After calling your function, the tree should look like:

         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \  / \
    4->5->6->7 -> NULL

[解题思路]

对二叉树使用层序遍历，层序遍历过程使用两个queue来分层

/**
 * Definition for binary tree with next pointer.
 * public class TreeLinkNode {
 *     int val;
 *     TreeLinkNode left, right, next;
 *     TreeLinkNode(int x) { val = x; }
 * }
 */
public class Solution {
    public void connect(TreeLinkNode root) {
        // Start typing your Java solution below
        // DO NOT write main() function
        if(root == null){
            return;
        }
        
        levelOrderTraverse(root);
    }
    
    public void levelOrderTraverse(TreeLinkNode node){
        LinkedList<TreeLinkNode> first = new LinkedList<TreeLinkNode>();
        LinkedList<TreeLinkNode> second = new LinkedList<TreeLinkNode>();
        first.add(node);
        while(!first.isEmpty()){
            TreeLinkNode tmp = first.poll();
            if(tmp.left != null){
                second.add(tmp.left);
            }
            if(tmp.right != null){
                second.add(tmp.right);
            }
            if(first.isEmpty()){
                tmp.next = null;
                first.addAll(second);
                second.clear();
            } else {
                tmp.next = first.peek();
            }
        }
    }
}

 updated

由于题目中要求只能使用constant space，所以上面使用两个queue做法是不对的

         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \  / \
    4->5->6->7 -> NULL

这题主要的难点在于处理节点5的next，因为5的next要指向它的sibling，但在5这层是没法获取到3的引用

解决方法是：由于2所在那层已经建立好next链接，所以只需由2即可得到3的引用，继而得到3的left节点

public void connect(TreeLinkNode root) {
        // Start typing your Java solution below
        // DO NOT write main() function
        if(root == null){
            return;
        }
        if(root.left != null){
            root.left.next = root.right;
        }
        if(root.right != null){
            root.right.next = (root.next != null) ? root.next.left : null;
        }
        
        connect(root.left);
        connect(root.right);
    }